nCaCO3=\(\dfrac{50}{100}\)=0,5(mol)
PTHH: CaCO3+2HCl→CaCl2+H2O+CO2↑
Theo PT ta có: nCaCO3=nCaCl2=nCO2=0,5(mol)
Theo PT ta có: nHCl=0,5.21=1(mol)
⇒mddHCl=\(\dfrac{1.36,5}{20\%}\)=182,5(g)
⇒mddsau−pư=mCaCO3+mHCl−mCO2
⇔mddHCl=50+182,5−22=210,5(g)
⇒C%CaCl2=\(\dfrac{0,5.111}{210,5}\).100%≈26,37%

nCaCO3 = 50/100 = 0,5 (mol(

PTHH: CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

Mol: 0,5 ---> 0,1 ---> 0,5 ---> 0,5 ---> 0,5

mHCl = 1 . 36,5 = 36,5 (g)

mddHCl = 36,5/20% = 182,5 (g)

mCO2 = 0,5 . 44 = 22 (g)

mdd (sau p/ư) = 50 + 182,5 - 22 = 210,5 (g)

mCaCl2 = 0,5 . 111 = 55,5 

C%CaCl2 = 55,5/210,5 = 26,36%

\(n_{NaOH}=\dfrac{200\cdot2\%}{40}=0.1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{50\cdot49\%}{98}=0.25\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(0.1...............0.05...........0.05\)

\(n_{Na_2SO_4}=0.05\left(mol\right)\)

\(n_{H_2SO_4\left(dư\right)}=0.25-0.05=0.2\left(mol\right)\)

\(V_{dd}=\dfrac{200}{1}+\dfrac{50}{1.05}=247.6\left(ml\right)=0.2476\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0.05\cdot2}{0.2476}=0.4\left(M\right)\)

\(\left[H^+\right]=\dfrac{0.2\cdot2}{0.2476}=1.6\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.2}{0.2476}=1\left(M\right)\)

$n_{NaOH} = \dfrac{200.2\%}{40} = 0,1(mol)$
$n_{H_2SO_4} = \dfrac{50.49\%}{98} = 0,25(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} : 2 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư

$n_{H_2SO_4\ dư} = 0,25 - 0,1.0,5 = 0,2(mol)$

$n_{H^+\ dư} = 0,2.2 = 0,4(mol)$

Sau phản ứng :

$V_{dd} = \dfrac{200}{1} + 50.1,05 = 252,5(ml) = 0,2525(lít)$

Bảo toàn Na, S ta có : 

$[Na^+] = \dfrac{0,1}{0,2525} = 0,4M$
$[SO_4^{2-}] = \dfrac{0,25}{0,2525} = 0,99M$
$[H^+] = \dfrac{0,4}{0,2525} = 1,58M$

\(n_{OH^-}=\dfrac{200.2\%}{40}=0,1\left(mol\right);n_{H^+}=2.\dfrac{50.49\%}{98}=0,5\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

0,5.........0,1

=> Sau phản ứng H+ dư

\(n_{H^+\left(dư\right)}=0,5-0,1=0,4\left(mol\right)\)

Dung dịch sau phản ứng gồm các ion : Na ⁺ ,H⁺ dư, SO4^2-

V=\(\dfrac{200}{1}+\dfrac{50}{1,05}=247,6\left(ml\right)=0,2476\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0,1}{0,2476}=0,4M\)

 \(\left[H^+_{dư}\right]=\dfrac{0,4}{0,2476}=0,8M\)

\(\left[SO_4^{2-}\right]=\dfrac{0,25}{0,2476}=1M\)

\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)

\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)

\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)

m dd = D.V = 1,206.200 = 241,2(gam)

C% CuSO4 = 50/241,2  .100% = 20,73%

\(n_{H^+}=n_{HCl}=0,03mol\)

\(n_{OH^-}=2n_{Ba\left(OH\right)_2}=0,07mol\)

          \(H^+\)     +      \(OH^-\)      \(\rightarrow\)      \(H_2O\)

bđ     0,03              0,07

pư     0,03              0,03                 0,03

kt         0                 0,04                 0,03

\(\Rightarrow\)\(\left[OH^-\right]=\dfrac{n_{sau}}{V}=\dfrac{0,04}{1}=0,04M\)

a) GS có x mol HCl

\(\Rightarrow m_{HCl}\)mHClHCl=36,5x

\(\Rightarrow m_{dd_{HCl}}\)=36,5x/37%=98,65x

\(\Rightarrow V_{dd}=\frac{m_{dd}}{D}\)=\(\frac{98,65x}{1,19}\)=82,9x (ml)

\(\Rightarrow CM_{dd_{HCl}}\)=x/0,0829x=12M

b) GS có x mol HCl

\(\Rightarrow m_{HCl}\)=36,5x

\(V_{dd_{HCl}}\)=x/10,81 lít

\(\Rightarrow m_{dd_{HCl}}\)=1/10,81.1000.1,17.x=108,233x

\(\Rightarrow\)C%=36,5/108,233.100%=33,72%

GS có x mol HClHCl

=>mHClHCl=36,5x

=>mdd HClHCl=36,5x/37%=98,65x

=>Vdd=mdd/D=98,65x/1,19=82,9x (ml)

=>CM dd HClHCl=x/0,0829x=12M

b) GS có x mol HClHCl

=>mHClHCl=36,5x

Vdd HClHCl=x/10,81 lít

=>mdd HClHCl=1/10,81.1000.1,17.x=108,233x

=>C%=36,5/108,233.100%=33,72%

a, \(\left\{{}\begin{matrix}n_{Ba^{2+}}=4.10^{-3}\left(mol\right)\\n_{Na^+}=3.10^{-3}\left(mol\right)\\n_{OH^-}=0,011\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\dfrac{4.10^{-3}}{0,2+0,3}=0,008M\\\left[Na^+\right]=\dfrac{3.10^{-3}}{0,2+0,3}=0,006M\\\left[OH^-\right]=\dfrac{0,011}{0,2+0,3}=0,022M\end{matrix}\right.\)

b, Để trung hòa dung dịch A thì:

\(n_{H^+}=n_{OH^-}\)

\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)

\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)

cần lời giải chi tiết ạ

\(n_{CaCO_3}=\dfrac{50}{100}=0,5\left(mol\right)\)
PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
Theo PT ta có: \(n_{CaCO_3}=n_{CaCl_2}=n_{CO_2}=0,5\left(mol\right)\)
Theo PT ta có: \(n_{HCl}=\dfrac{0,5.2}{1}=1\left(mol\right)\)
\(\Rightarrow mdd_{HCl}=\dfrac{1.36,5}{20\%}=182,5\left(g\right)\)
\(\Rightarrow mdd_{sau-pư}=m_{CaCO_3}+m_{HCl}-m_{CO_2}\)
\(\Leftrightarrow mdd_{HCl}=50+182,5-22=210,5\left(g\right)\)
\(\Rightarrow C\%_{CaCl_2}=\dfrac{mct}{mdd}.100\%=\dfrac{0,5.111}{210,5}.100\%\approx26,37\%\)

\(CuCl_2+H_2S\rightarrow CuS+2HCl\)

\(m_{dd_{CuCl2}}=3,38.50=169\left(g\right)\)

\(n_{CuCl2}=\frac{169.20\%}{135}=0,25\left(mol\right)\)

\(n_{H2S}=\frac{50.20,4\%}{34}=0,3\left(mol\right)\)

Suy ra H2S dư

\(n_{CuS}=n_{CuCl2}=0,25\left(mol\right)\)

\(n_{HCl}=0,25.2=0,5\left(mol\right)\)

\(m_{dd_{Spu}}=169+50-0,25.96=195\left(g\right)\)

\(C\%_{HCl}=\frac{0,5.36,5}{195}.100\%=9,36\%\)